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Chapter 2: Problem 61

A person who is properly restrained by an over-the-shoulder seat belt has a good chance of surviving a car collision if the deceleration does not exceed 30 "\(g\)'s" (1.00 \(g =\) 9.80 m/s\(^2\)). Assuming uniform deceleration at 30 \(g\)'s, calculate the distance over which the front end of the car must be designed to collapse if a crash brings the car to rest from 95 km/h.

Short Answer

Expert verified
The front end of the car must collapse over approximately 1.18 meters.

Step by step solution

01

Given Variables and Conversion

We begin by identifying the given data and convert as necessary. The initial speed of the car, \( v_i \), is 95 km/h. We need to convert this to meters per second:\[v_i = 95 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 26.39 \text{ m/s}\]The deceleration, \( a \), is given as 30 \( g \)s and needs to be converted into meters per second squared:\[a = 30 \times 9.8 \text{ m/s}^2 = 294 \text{ m/s}^2.\]
02

Applying the Kinematic Equation

We need to find the stopping distance, \( d \). Using the kinematic equation:\[ v_f^2 = v_i^2 + 2ad \]where \(v_f\) is the final velocity which is 0 m/s (since the car comes to a stop), we can rearrange the formula to solve for \(d\):\[ 0 = (26.39 \text{ m/s})^2 + 2(-294 \text{ m/s}^2)d \]
03

Solving for Distance

Rearrange the formula from the previous step to solve for \(d\):\[2 \times 294 \times d = (26.39)^2 \d = \frac{(26.39)^2}{2 \times 294} \]Calculate \(d\):\[d = \frac{696.47}{588} \approx 1.18 \text{ m}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Deceleration
Deceleration is the rate at which an object slows down. In kinematics, it is often represented in terms of acceleration, but with a negative value, indicating a reduction in speed.

Deceleration is crucial in scenarios like vehicle crashes, as it determines how quickly a vehicle can come to a stop. It is measured in meters per second squared (m/s虏). For this specific problem, deceleration is given in "g-forces" where 1g equals 9.80 m/s虏.
  • To calculate deceleration in standard units, you multiply the given g-force by 9.80 m/s虏.
  • In our scenario, with a deceleration of 30 g's, the actual deceleration in meters per second squared becomes: 30 x 9.80 m/s虏 = 294 m/s虏.
Sufficiently calculating deceleration helps us understand the force exerted on passengers and the vehicle structure during a collision.
Stopping Distance
Stopping distance is the length a vehicle travels from the point the brakes are applied until it comes to a complete stop.

This is especially important in safety engineering when designing car features to ensure passenger safety during a crash. In our exercise, we calculated the stopping distance to understand how much space is required to safely slow the car down in an accident.
  • The stopping distance depends on the initial speed of the vehicle, the applied deceleration, and the condition of the road.
  • The kinematic equation used: \[ v_f^2 = v_i^2 + 2ad \]helps calculate this distance by setting the final velocity \( v_f \) to 0, as the vehicle comes to a stop.
  • By solving it, we found that the car must collapse over a distance of approximately 1.18 meters to absorb the impact safely.
The stopping distance is a vital aspect of vehicle design, influencing features like the crumple zones that absorb kinetic energy during a collision.
Uniform Acceleration
Uniform acceleration occurs when an object's acceleration is constant, resulting in a predictable change in its velocity over time.

This concept is readily applied in kinematics problems involving cars and other moving objects, as in our exercise. It is an assumption that simplifies the problem-solving process because it allows us to use constant values in the equations.
  • In the context of deceleration in a car crash, uniform acceleration implies that the car loses velocity at a steady rate until it comes to rest.
  • By assuming uniform acceleration, engineers can predict stopping distances more accurately, ensuring safer vehicle designs.
  • In the given problem, understanding uniform acceleration allows us to compute the stopping distance by using a steady deceleration rate of 294 m/s虏, as derived from 30 g's.
This predictability of motion with uniform acceleration is pivotal in designing safety mechanisms in vehicles, like crumple zones, designed to deform at a predictable rate to maximize safety during sudden stops.

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Most popular questions from this chapter

(III) An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road. If the train's speed is 75 km/h, how long does it take the car to pass it, and how far will the car have traveled in this time? See Fig. 2-36. What are the results if the car and train are traveling in opposite directions?

(II) A ball player catches a ball 3.4 s after throwing it vertically upward. With what speed did he throw it, and what height did it reach?

A car is behind a truck going 18 m/s on the highway. The car's driver looks for an opportunity to pass, guessing that his car can accelerate at 0.60 m/s\(^2\) and that he has to cover the 20-m length of the truck, plus 10-m extra space at the rear of the truck and 10 m more at the front of it. In the oncoming lane, he sees a car approaching, probably at the speed limit, 25 m/s (55 mph). He estimates that the car is about 500 m away. Should he attempt the pass? Give details.

A bicyclist in the Tour de France crests a mountain pass as he moves at 15 km/h. At the bottom, 4.0 km farther, his speed is 65 km/h. Estimate his average acceleration (in m/s\(^2\)) while riding down the mountain.

Suppose a car manufacturer tested its cars for front-end collisions by hauling them up on a crane and dropping them from a certain height. (\(a\)) Show that the speed just before a car hits the ground, after falling from rest a vertical distance \(H\), is given by \(\sqrt{ 2gH }\) . What height corresponds to a collision at (\(b\)) 35 km/h? (\(c\)) 95 km/h?
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